question_answer

A stone of mass 1 kg tied to a light inextensible 10 string of length, L = \[\frac{10}{3}\]m is whirling in a circular 0 path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum is 4 and if g is taken to be 10\[\text{m}/{{\text{s}}^{\text{2}}}\], then speed of stone at the highest point of the circle is

A)  20 m / s              

B)         \[10\sqrt{3}m/s\]

C)  \[50\sqrt{2}m/s\]            

D)         10 m / s

Correct Answer: C

Solution :

 Here, \[r=L=\frac{10}{3}m\]             \[{{T}_{\max }}=mg+\frac{m{{v}_{2}}^{2}}{r}\] and       \[{{T}_{\min }}=-mg+\frac{m{{v}_{1}}^{2}}{r}\] Given:   \[\frac{{{T}_{\max }}}{{{T}_{\min }}}=4\] \[\Rightarrow \]   \[{{T}_{\max }}=4{{T}_{\min }}\] \[\therefore \]      \[mg+\frac{m{{v}_{2}}^{2}}{r}=-4mg+\frac{4m{{v}_{1}}^{2}}{r}\] \[\Rightarrow \]   \[5g+\frac{{{v}_{2}}^{2}}{r}=\frac{4{{v}_{1}}^{2}}{r}\] \[\Rightarrow \]   \[5\,\,g\,\,r+{{v}_{2}}^{2}=4{{v}_{1}}^{2}\] \[\Rightarrow \]   \[5g\times \frac{10}{3}+{{v}_{1}}^{2}+2g\times \frac{10}{3}\times 2=4{{v}_{1}}^{2}\]                                     \[(\because \,\,{{v}_{2}}^{2}={{v}_{1}}^{2}-2gh)\] \[\Rightarrow \]   \[\frac{50g}{3}+\frac{40g}{3}=3{{v}_{1}}^{2}\] \[\Rightarrow \]   \[\frac{90g}{3}=3{{v}_{1}}^{2}\] \[\Rightarrow \]   \[10\times 10={{v}_{1}}^{2}\] \[(\because \,\,g=10\,\,m/{{s}^{2}})\] \[\therefore \]      \[{{v}_{1}}=\mathbf{10}\,\,\mathbf{m/s}\]